The given limit problem is:

$\lim_{x \to 1} \frac{4x + 3}{x - 1}$

Solution:

1. Substitute $x = 1$ directly:

$\frac{4(1) + 3}{1 - 1} = \frac{4 + 3}{0} = \frac{7}{0}$

Since the denominator becomes zero, this limit indicates a vertical asymptote, implying that the expression tends towards infinity or negative infinity. To determine this precisely, we need to analyze the behavior of the function as $x \to 1^+$ (from the right) and $x \to 1^-$ (from the left).

2. As $x \to 1^+$ (from the right):

For values of $x$ slightly greater than 1, $x - 1$ is positive. Therefore,

$\frac{4x + 3}{x - 1} \to +\infty$

3. As $x \to 1^-$ (from the left):

For values of $x$ slightly less than 1, $x - 1$ is negative. Therefore,

$\frac{4x + 3}{x - 1} \to -\infty$

Conclusion:

The limit does not exist as it approaches different infinities from the left and right:

$\lim_{x \to 1} \frac{4x + 3}{x - 1} = \text{DNE} \quad (\text{Does Not Exist}).$

Would you like any further details or clarifications?

Here are 5 related questions for further exploration:

1. What is the general approach to solving limits involving division by zero?
2. How do you determine if a limit approaches $+\infty$ or $-\infty$?
3. Can this limit be evaluated using L'Hopital's Rule?
4. What is a vertical asymptote, and how is it related to limits?
5. How does one interpret one-sided limits in real-world scenarios?

Tip: When dealing with limits that result in a zero denominator, always check the behavior from both the left and right sides of the point of interest.